高精度
vector
{
vector
long long t = v[0];
for (int i = 1; i <= v.size(); i++)
{
ans.push_back(t / b);
t %= b;
t *= 10;
t += v[i];
}
while (ans.size() > 1 && ans.front() == 0)
ans.erase(ans.begin());
return ans;
}
本文来自博客园,作者:多巴胺不耐受仿生人,转载请注明原文链接:https://www.cnblogs.com/VoidCoderTF/articles/15435488.html

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